To solve a trig equation in a given interval, find the base angle from your calculator using the positive value, then use the graph to place every solution in the interval. Equations that rearrange into a quadratic give two separate equations to solve, and equations with a transformed angle such as sin(2x − 30°) need the interval transformed before you solve.
The method at a glance
- sin x = k: base angle α, solutions α and 180° − α (then add 360° as needed)
- cos x = k: base angle α, solutions α and 360° − α
- tan x = k: base angle α, solutions α, α + 180°, …
Transformed angle, such as sin(2x − 30°) = k
- Let u = 2x − 30°.
- Transform the interval for u: double, then subtract 30°.
- Solve for u across the whole new interval.
- Convert back: x = (u + 30°) / 2.
- Check each x is in the original interval.
How do you solve a trig equation in a given interval?
Two steps. First, find the base angle from the calculator using the positive value of k. Second, use the graph to place every solution in the interval: the base angle gives the size, and the graph shows where the curve actually reaches that height.
For sine and cosine, a horizontal line crosses each 360° cycle twice. For tangent, the period is only 180°, so the second solution is the first plus 180°. Take the base angle from the positive value; the graph, not the calculator, tells you where negative values sit.
Equations that rearrange into a quadratic
Many equations become a quadratic in one trig function. Once it is factorised, each bracket gives its own trig equation, and each must be solved across the whole interval.
Three things to watch. A root such as cos x = −2 is impossible and must be rejected. Square-rooting needs a ±. And a value like sin x = −1, where the curve only touches the line, still gives one solution. The most common loss of marks is solving one bracket and stopping.
Transformed angles: shifting the interval
For an equation like sin(x + 30°) = 0.5, you are not solving for x directly. Substitute u = x + 30°, and move the interval with it: 0° ≤ x < 360° becomes 30° ≤ u < 390°. Solve for u in that interval, then subtract 30° to get x.
Shifting the interval is the step students skip, and it is exactly where solutions are lost or wrongly included.
Transformed angles: scaling the interval
For sin 2x = k, substitute u = 2x and multiply both ends of the interval by 2: 0° ≤ x < 360° becomes 0° ≤ u < 720°. A doubled interval contains twice as many cycles, so expect twice as many solutions. Convert back with x = u / 2.
For the hardest form, such as sin(2x − 30°), apply the operations in the same order as the expression: double the ends, then subtract 30°. A half-open interval such as u < 690° excludes the endpoint itself.
“Show that”, then hence solve
The full exam question uses an identity in part (a) to turn the equation into a quadratic, then says “hence solve” in part (b). Part (b) usually carries more marks, and it is where the interval work happens. If part (a) gives you the quadratic, use it in part (b) even if your own working in (a) went wrong; you can still earn the (b) marks.
Worked examples
Example 1. Solve sin x = 0.6 for 0° ≤ x < 360°, to 1 d.p.
- Base angle: sin⁻¹(0.6) = 36.9°.
- The sine curve is above the axis from 0° to 180°, so both solutions are there.
- By symmetry about 90°, the second is 180° − 36.9° = 143.1°.
Answer: x = 36.9°, 143.1°.
Marks: M1 for the base angle, A1 for both solutions.
Example 2. Solve 2 cos²x + 3 cos x − 2 = 0 for 0° ≤ x < 360°.
- Factorise: (2 cos x − 1)(cos x + 2) = 0.
- So cos x = ½ or cos x = −2.
- Reject cos x = −2, since cos x always lies between −1 and 1.
- cos x = ½: base angle 60°, so x = 60° and x = 360° − 60° = 300°.
Answer: x = 60°, 300°.
Marks: M1 for factorising, B1 for rejecting cos x = −2, A1 for both solutions.
Example 3. Solve sin(x + 30°) = 0.5 for 0° ≤ x < 360°.
- Let u = x + 30°, so sin u = 0.5.
- Shift the interval: 30° ≤ u < 390°.
- Solutions for u: 30° and 180° − 30° = 150°. The next, 390°, is not included because u < 390°.
- Convert back: x = u − 30°, giving 0° and 120°.
Answer: x = 0°, 120°.
Marks: B1 for the shifted interval, M1 for solving for u, A1 for both values of x.
Example 4. Solve sin(2x − 30°) = √3/2 for 0° ≤ x < 360°.
- Let u = 2x − 30°, so sin u = √3/2.
- Transform the interval: doubling gives 0° ≤ 2x < 720°, then subtracting 30° gives −30° ≤ u < 690°.
- Base angle 60°. First cycle: u = 60°, 120°. Second cycle: u = 420°, 480°.
- Convert back with x = (u + 30°) / 2: x = 45°, 75°, 225°, 255°.
Answer: x = 45°, 75°, 225°, 255°.
Marks: B1 for the transformed interval, M1 for all values of u, A1 for all values of x.
Common mistakes
- Taking the base angle from a negative value. Find the base angle from the positive value, then use the graph to place the solutions.
- Solving only one bracket. Each factor of a quadratic gives its own set of solutions.
- Not transforming the interval. For sin(x + 30°) or sin 2x, change the interval for u before solving, or solutions will be missed.
- Including an endpoint the interval excludes. With u < 390°, the value u = 390° is not a solution.
- Missing a touching root. sin x = −1 or cos x = 1 gives exactly one solution in each cycle, but it still counts.
How trig equations are examined
Solving trig equations is Pure 2 content in Edexcel International A-Level (paper WMA12), in the trigonometric identities and equations chapter. In Edexcel A-Level (9MA0) it is AS (Year 1) Pure. You will usually be asked to:
- solve a basic equation in sin, cos or tan over an interval, giving all solutions
- show that an equation becomes a quadratic, then hence solve it
- solve an equation with a transformed angle such as sin(2x − 30°)
- explain why a root is rejected
Find full questions with mark schemes in the Pure 2 trig identities and equations exam questions, or work through the trig identities and equations worksheet. For unlimited extra equations, use the A-Level trigonometry revision engine.
Before this course, make sure you are confident with sin²θ + cos²θ = 1 and tan θ = sin θ / cos θ. In Pure 3, further trig equations add double angles and sec, cosec and cot.
Frequently asked questions
How do I find all solutions of a trig equation?
Find the base angle, then use the graph and the period to list every solution in the interval: for sine and cosine there are usually two per 360°, and for tangent one per 180°.
Why does sin 2x = k have more solutions than sin x = k?
Because u = 2x runs over an interval twice as long, so the curve completes twice as many cycles and crosses the line twice as often.
What does it mean to reject a root?
A quadratic in sin x or cos x can produce a value outside −1 to 1, such as cos x = −2. No angle has that cosine, so you discard it and say why.